# What is the integral of this gaussian [duplicate]

I want to know what is the following integral

$\int e^{-(y-\mu)^T \Lambda(y-\mu) } dy$ I am trying to see the properties of gaussian integral but I couldn't find anything for this one. Any help guys?

I want to know how given

where Z(x) is the partition function

This is the paper

If the integration that you guys have given is correct there should be $|\Lambda|^{1/2}$ at the end instead of just $|\Lambda|$ in $\frac{1}{Z(x)}$. I doubt that there is a type in the paper. So I must be missing something

Can anyone explain why that guy didn't have $|\Lambda|^{1/2}$ but $|\Lambda|$ instead???

## marked as duplicate by Did, user61527, Stefan4024, Dennis Gulko, Davide GiraudoOct 15 '13 at 17:09

• Sorry but this is not the way the site is supposed to function: you are not supposed to post an embryo of question then to wait for answers then to add some crucial information to the question (before maybe some more rounds of the same?). Three more things: notwithstanding your "doubts", $|\Lambda|$ in the paper is obviously a typo (and I mean, obviously); yes "the integration that [I] have given is correct", thanks; I delete my answer. – Did Oct 15 '13 at 16:03
For the integral to exist, $\Lambda$ has to be positive definite. If $\Lambda$ is positive definite, then we can decompose $\Lambda$ as $\Lambda = R^TR$ (for instance, Cholesky decomposition). We then have $$I = \int_{\mathbb{R}^n} \exp \left(-(y-\mu)^T R^T R (y- \mu)\right)dy$$ Let $x = R(y -\mu)$. We then have $dx = \det(R) dy$. Hence, \begin{align} I & = \int_{\mathbb{R}^n} \exp \left(-x^T x\right)\dfrac{dx}{\det(R)} = \dfrac1{\det(R)} \prod_{k=1}^n \int_{-\infty}^{\infty}\exp(-x^2)dx\\ & = \dfrac1{\det(R)} \prod_{k=1}^n \sqrt{\pi} = \dfrac{\pi^{n/2}}{\det(R)} = \sqrt{\dfrac{\pi^n}{\det(\Lambda)}} \end{align}